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Why we can do z=a+bi assumption? Any prove show what we assume is correct and how do we know the answer come out is correct?

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- Thread starter Outrageous
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- #1

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Why we can do z=a+bi assumption? Any prove show what we assume is correct and how do we know the answer come out is correct?

- #2

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Sorry, can you tell what is the subtopic of my question in the reference given. Thanks

- #4

HallsofIvy

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I'm not sure what you mean by "z= a+ bi assumption". Typically that is NOT an "assumption", it is one way of

Why we can do z=a+bi assumption? Any prove show what we assume is correct and how do we know the answer come out is correct?

A more formal definition, from which you can then derive "z= a+ bi" is this:

The complex numbers is the set of all

One can also show that [itex](0, 1)^2= -1[/itex]: (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+1(0))= (-1, 0) which, as above, we identify with the real number -1. We

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One can also show that [itex](0, 1)^2= -1[/itex]: (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+1(0))= (-1, 0) which, as above, we identify with the real number -1. Wedefine"i= (0, 1)" so that [itex]i^2= -1[/itex]. And, then, we can write (a, b)= (a, 0)+ (0, b)= (a, 0)+ (b, 0)(0, 1)= a+ bi.

Thank you.

Can I say all the number actually is the combination of real and imaginary number.

As you say" We can "identify" complex numbers of the form (a, 0) with the real number, a"

- #6

HallsofIvy

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_{1+ b1i and the same z= a2+ b2i, we must have a1= a2 and b1= b2. Now, in this problem, attempting to find (2+ 6i)1/3, they are suggesting that you write the answer as "a+ bi", find (a+ bi)3 and set the "real" and "imaginary" parts equal. For example, (a+ bi)2= (a+ bi)(a+ bi)= a2+ 2abi+ b2i2= (a2- b2+ 3abi. If we wanted to find (1+ i)1/2, we could now set s2- b2= 1 and ab= 1, giving two equations to solve for a and b.}

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